When we experience positive "g forces", it is as if we have become...

A 6.00-m long string sustains a three-loop standing wave pattern as shown. The wave speed is 2.00 × 102 m/s.What is the lowest p

KATRIN_1 [288]

Given:

The length of the string is l = 6 m

The speed of the wave is

$v=2\times10^2\text{ m/s}$

Required: Lowest possible frequency for the standing wave.

Explanation:

The lowest possible frequency is the fundamental frequency.

The fundamental frequency can be calculated by the formula

$f=\frac{v}{2l}$

On substituting the values, the fundamental frequency will be

$\begin{gathered} f=\frac{2\times10^2}{2\times6} \\ =16.67\text{ Hz} \end{gathered}$

Final Answer: The lowest possible frequency for standing waves on this string is 16.67 Hz

4 0

2 years ago

A small spaceship with a mass of only 2.8 ✕ 103 kg (including an astronaut) is drifting in outer space with negligible gravitati

Luda [366]

Answer:

962.14m/s

Explanation:

Data obtained from the question include:

m (mass) = 2.8x10^3 kg

P (power) = 30 kW = 30 x 1000 = 30000W

V (velocity) =?

t (time) = 1 day

There are 24 hours in a day.

t (time) = 1 day = 24 hours

We need to covert 24 hours to seconds. This is illustrated below:

There are 60 minutes in 1 hour and 60 seconds in 1 minutes.

Therefore, 24hours = 24 x 60 x 60 = 86400 seconds.

t (time) = 1 day = 24 hours = 86400 seconds

Power is related to velocity according to equation:

Power = force x Velocity

P = F x v (1)

Recall Force (F) = Mass (m) x a (acceleration) i.e F = ma

Substituting the value of F into equation 1, we have:

P = F x v

P = ma x v

P = m x a x v (2)

But: acceleration (a) = Velocity(v)/time(t) i.e a = v/t

Substituting the value of a into equation 2, have:

P = m x a x v

P = m x v/t x v

P = (m x v^2)/ t

Now, with this equation

P = (m x v^2)/ t, we can obtain the speed of the spaceship as follow:

P = (m x v^2)/ t

30000 = (2.8x10^3 x v^2) /86400

Cross multiply to express in linear form

2.8x10^3 x v^2 = 30000 x 86400

Divide both side by 2.8x10^3

v^2 = (30000 x 86400)/ 2.8x10^3

v^2 = 925714.2857

Take the square root of both side

v = √(925714.2857)

v = 962.14m/s

Therefore, the speed of the spaceship is 962.14m/s

3 0

3 years ago

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A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.2

slavikrds [6]

The Electric current is 1.11* 10^{-4}A

Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

To learn more about electric current please visit -brainly.com/question/12791045
#SPJ1

7 0

2 years ago

Read 2 more answers

Elements on the periodic table are grouped by their

vodka [1.7K]

Similar chemical behavior. All the members of a group of elements have the same number of valence electrons and similar chemical properties.

5 0

4 years ago

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Two 2.0-cm-diameter insulating spheres have a 6.60 cm space between them. One sphere is charged to + 76.0 nC , the other to - 30

e-lub [12.9K]

Answer:

$5.2\times 10^5N/C$

Explanation:

Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.

This can be easily seen if we use Gauss's law, $\int{E} \, dA=\frac{Q_{enclosed}}{\epsilon_o}$

We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

$E=\frac{1}{4\pi r^2} \frac{Q_{enclosed}}{\epsilon_o}$

which is the same as that of a point charge.

In our problem the charges being of opposite signs, the electric field will add up. Therefore,

$E_{total}=\frac{1}{4\pi\epsilon_o}\frac{q_1+q_2}{r^2}= (9\times10^9) \frac{(76+30)\times10^{-9}}{((1+3.3)\times10^{-2})^2}N/C =5.2\times10^5N/C$

where, $r$ = distance between the center of one sphere to the midpoint (between the 2 spheres)

8 0

3 years ago